To solve this, we will use the formula for compound interest:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Where:

• $A$ is the amount of money in the account after $t$ years ($75,000$ in this case),
• $P$ is the principal amount (what we are solving for),
• $r$ is the annual percentage rate as a decimal ($4\% = 0.04$),
• $n$ is the number of times the interest is compounded per year ($365$ for daily compounding),
• $t$ is the number of years the money is invested ($14$).

Rearranging to solve for $P$:

$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$

Substituting the given values:

$P = \frac{75,000}{\left(1 + \frac{0.04}{365}\right)^{365 \cdot 14}}$

Calculation:

1. Calculate the daily interest rate: $\frac{0.04}{365} = 0.000109589$.
2. Add 1 to the rate: $1 + 0.000109589 = 1.000109589$.
3. Raise to the power $365 \cdot 14 = 5110$: $(1.000109589)^{5110}$.
4. Divide $75,000$ by the result of Step 3.

Let me compute this.# Given values A = 75000 # Future value r = 0.04 # Annual percentage rate (4%) n = 365 # Compounded daily t = 14 # Time in years

Calculate the principal (P)

daily_rate = r / n compound_factor = (1 + daily_rate) ** (n * t) P = A / compound_factor

P42841.994276135442841.9942761354successThe initial deposit needed is approximately $42,841.99.

If you invest this amount now at an APR of 4% compounded daily, it will grow to $75,000 in 14 years.

Would you like further clarification or additional calculations? Here are some related questions to consider:

1. How does the calculation change if the interest is compounded monthly instead of daily?
2. What is the total interest earned over the 14 years?
3. How much would you need to invest with a 6% APR compounded daily?
4. What if you wanted to reach $75,000 in 10 years instead of 14 years?
5. How does the frequency of compounding impact the final amount?

Tip: Compounding frequency plays a significant role in determining the growth of investments. The more frequent the compounding, the greater the final amount.